Tuesday, April 13, 2010

Fox 274

2 comments:

  1. Let the circle be:(x-1/2)^2+(y-r)^2=r^2
    The parabola:y=8x-8x^2. These two equations can be solved simultaneously and the discriminant set to 0 to obtain points of tangency. This gives us: 256r^2+32r -63=0 which, in turn, yields r = 7/16 as the only admissible root.
    Ajit

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  2. Yu:
    Translate the parabola and the circle so that the y-axis is the axis of symmetry.
    y= -8(x^2 - 1/4), dy/dx = -16x.
    x^2 + (y-r)^2 = r^2,
    top-half is y=sqrt(r^2-x^2)+r, dy/dx = -x/sqrt(r^2-x^2).
    Let -8(x^2 - 1/4) = sqrt(r^2-x^2)+r
    and -16x = -x/sqrt(r^2-x^2).
    Solve for r = 7/16.

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